
smooth function
show that
$\displaystyle f(x,y,z)=\frac{1}{z^2+1}(x^2y^2,2xy,2z)$ defines a smooth function $\displaystyle f: S^{2}\to S^{2}$ where $\displaystyle S^{2}=\{p\in \mathbb{R}^3, p=1\}$ so I am guessing it is a unit sphere.
How do we proceed with such problem?
f is a smooth function if all partial derivatives of all possible orders are defined in all points of the domain of f right?
I started with computing these partial derivatives
$\displaystyle \frac{\partial f}{\partial x}=\frac{1}{z^2+1}(2x,2y,0)$
$\displaystyle \frac{\partial f}{\partial y}=\frac{1}{z^2+1}(2y,2x,0)$
$\displaystyle \frac{\partial f}{\partial z}=\frac{2(z^21)}{(z^2+1)^2}$
and these partial derivatives are all defined at any point from the domain of f
so I compute second order partial derivatives
$\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}$
$\displaystyle \frac{\partial^2 f}{\partial y^2}=\frac{2}{z^2+1}$
$\displaystyle \frac{\partial^2 f}{\partial z^2}=\frac{4z(z^23)}{(z^2+1)^3}$
and these 3 are also defined for all the points of the domain.
if I continue differentiating
$\displaystyle \frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0$ and
$\displaystyle \frac{\partial^3 f}{\partial z^3}=\frac{12(z^46z^2+1)}{(z^2+1)^4}$ which is defined for all points
What should I do next? can I conclude based on the calculations that f is certainly smooth?
Thanks in advance

Re: smooth function
f is obviously smooth on R^3. So what you need to show is, when f is restricted to S^2, its image is in S^2.
That is, to show that [(x^2y^2)^2 + (2xy)^2 + (2z)^2]/(1+z^2)^2 = 1, given that x^2+y^2+z^2=1.
Which is easy to show.

Re: smooth function
so showing all the computations above were not necessary if I am supposed to restrict myself to $\displaystyle S^{2}$ right?
But I still don't get how to show $\displaystyle \Biggl[(x^2y^2)^2+(2xy)^2+(2z)^2\Biggr]\frac{1}{(z^2+1)^2}=1$ how does this imply that f will be smooth on the sphere???

Re: smooth function
$\displaystyle (x^2y^2)^2 + (2xy)^2 + (2z)^2=(x^2+y^2)^2 + (2z)^2$
=$\displaystyle (1z^2)^2+(2z)^2=(1+z^2)^2$
This shows that $\displaystyle f_{S^2}$ indeed maps to $\displaystyle S^2$.
Then $\displaystyle f$ is smooth because "the restriction of a smooth function on a smooth submanifold is smooth".
To prove the above statement, let $\displaystyle i: S^2 \rightarrow R^3$ is the smooth embedding of $\displaystyle S^2$ in $\displaystyle R^3$,
$\displaystyle f : R^3 \rightarrow S^2 $ is a smooth map on $\displaystyle R^3$, then $\displaystyle f_{S^2} = f \circ i$ is smooth on $\displaystyle S^2$.